details

property	value
status	complete
benchmark	4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.91773 seconds
cpu usage	4.41054
user time	4.19405
system time	0.216491
max virtual memory	1.8412216E7
max residence set size	319376.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 0 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, f(a, y)) -> f(f(y, f(f(a, a), a)), x) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, f(a, y)) -> F(f(y, f(f(a, a), a)), x) F(x, f(a, y)) -> F(y, f(f(a, a), a)) F(x, f(a, y)) -> F(f(a, a), a) F(x, f(a, y)) -> F(a, a) The TRS R consists of the following rules: f(x, f(a, y)) -> f(f(y, f(f(a, a), a)), x) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, f(a, y)) -> F(f(y, f(f(a, a), a)), x) The TRS R consists of the following rules: f(x, f(a, y)) -> f(f(y, f(f(a, a), a)), x) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, f(a, y)) -> F(f(y, f(f(a, a), a)), x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. F(x, f(a, y)) -> F(f(y, f(f(a, a), a)), x) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO,RATPOLO]: popout

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actions

all output return to TRS Standard